PHYS 3100
(Notes follow and parts taken from sources in bibliography)
Sinusoidal Sources
Circuits powered by current or voltage sources which oscillate sinusoidally are extremely
common. We’ve looked at the way resistors, capacitors, and inductors behave with no sources,
with step-function sources, and we’ve also looked at long-term (steady-state) DC behavior.
If the source never “settles down” to a constant value, we can expect the same from voltages and
currents in the circuit. From studying mechanical oscillators, you probably already know that 1)
they have a natural frequency that describes their motion when released from a state of nonzero
potential energy and 2) if they are driven by an external and time-varying force, their motion will
adopt the same frequency as that of the driving force rather than their natural frequency as time
goes on.
Your book writes sinusoidally varying voltages in the form
V ( t ) = Vm cos( ω t + φ
)
where Vm is the amplitude of the voltage, ω is the angular frequency of the driving source
(generally assumed to be constant) and φ is the phase angle which is used to match the
expression to the initial conditions (for example, if the voltage is at its minimum value at t = 0,
the phase angle is chosen so that Cos(0 + φ) = -1. In this case, φ = 180°.).
Frequently, we’ll want to know what DC voltage would be most nearly equivalent to a given AC
voltage. This is determined by finding the rms, or root-mean-squared value of the AC signal.
To do this, we just square the expression above for the voltage, integrate it over one full period,
divide by the length of that period, and take the square root of the result. We can write
Vrms =
1
T
∫
t0 + T
t0
Vm2 cos 2 ( ω t + φ ) dt
The integral of cos2 over one period is just ½, so we could write
Vrms =
Vm
2
In this way, we can see that a resistor connected across an alternating potential difference with
Vrms = 20 volts (for example) will dissipate the same power as it would if connected to a 20 V
battery. Similarly, we define an average or effective current as
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PHYS 3100
I rms =
Im
2
If we examine the behavior of a series RL circuit which has no current flowing in it at time t=0
(the instant a switch closes) and ask what happens after that time, we will still use Kirchoff’s
Voltage Law to get
L
dI
+ IR = Vm cos( ω t + φ
dt
)
which has, as a general solution,
I (t) =
− Vm
R + (ω L)
2
2
cos( φ − θ ) e − R t / L +
Vm
R + (ω L)
2
2
cos( ω t + φ − θ
)
In this expression, θ is defined by tan θ = ωL/R. The behavior of this solution (consisting of a
transient term added to a steady-state term) for a few different values of R, L, and φ is shown
below. For all plots, Vm = 1 volt.
In the first example below, we have R = 100 Ω, L = 5 H, ω = 20 rad/s, and φ = - π/2. The two
solutions are similar in magnitude for the first 0.1 second or so, but the transient term dies out
very shortly after that.
In the next pair of images, we have changed the phase from φ = -π/2 to φ = π/2, which only
changes our starting point.
2
PHYS 3100
Notice that in the third pair of figures (R = 10,000 Ω, L = 0.005 H, ω = 20 rad/s and φ = π/2) L is
so small and R is so large that the transient solution is essentially negligible compared to the
steady-state solution. This means that the exponential damping factor in the first part of the
general solution for the current very quickly extinguishes the transient response. Also, the larger
resistance more than makes up for the smaller inductance when looking at the magnitude of the
steady-state current, as it is much smaller now than in the previous examples.
Phasors and Complex Numbers
We’ve seen that the voltage across an inductor depends on the time derivative of the current
through it, and the voltage across a capacitor can be thought of as the time integral of the
current to it. If the voltage or current driving a circuit changes sinusoidally, the time derivative
and time integral of that voltage or current will be out of phase with the driving force. The
voltage and current in a resistor are always in phase (since V = I R has no time derivatives or
integrals).
When resistors and capacitors and/or inductors are all present in a circuit, the way they combine
to affect the current can be tricky to determine. The capacitor and inductor are known as reactive
elements, while the resistor is (surprise) a resistive circuit element. Finding the reactive and
resistive contributions to a circuit is very similar to the process of two-dimensional vector
addition, where reactance is on one axis and resistance is on the other. We are used to writing
things like 6 iˆ − 5 ˆj when we mean the vector formed by moving 6 units in the positive x
3
PHYS 3100
direction and 5 units in the negative y direction, and we’re also familiar with writing the same
vector as a magnitude and direction (√61 and –39.8° in this case).
We can do the same thing to represent the relative contributions of resistance and reactance,
known as the phase angle. Any complex number of the form x + iy can be written in the form
Aeiθ where A is the amplitude of the number and θ is the phase. According to Euler, we could
also write the phase part of this as
e iθ = cos θ + i sin θ
Notice that there is no frequency information in this form. Also, note that we can extract either
the real or imaginary part of this expression to get cosθ or sinθ. Finally, keep in mind that some
textbooks (typically those for engineers) use j instead of i to represent √(-1). One other way to
write the complex number Aeiθ is A ∠θ°. Because the book uses cosines for these functions, if
you see something like 75 ∠-12°, that is equivalent to 75 cos(ω t-12°).
It takes a little practice (or a good calculator) to work with complex numbers, but they are very
important tools when dealing with reactive circuits. Since the voltage between the two terminals
of an AC generator alternates from its positive maximum to its negative minimum and back to
the maximum once each cycle, we can imagine the measured voltage as being like the shadow of
the hand of a large clock that is illuminated from the 12 o’clock position, as shown below:
12
9
12
3
9
6
3
6
As you can see, the length of the shadow will change periodically with the frequency of the
sweep. The shadow’s length corresponds to the voltage between the two terminals, and the
length of the clock hand is the amplitude of the oscillation. One difference between this model
and the phasor concept is that phasors (by convention, of course) rotate counter-clockwise. If we
are going to use the cosine representation, we will need the projection of the phasor on the real
axis rather than the imaginary one (which would give us the sine term). For an ordinary wall
outlet in the US, the magnitude of the phasor is about 170 V and it rotates at 60 revolutions per
second (meaning ω is about 377 rad/s).
Being comfortable with both ways of writing complex numbers is important, because if we’re
adding/subtracting two or more complex currents or voltages, it will be much easier to put them
in x + i y form and combine the real and imaginary parts separately. On the other hand, if we’re
trying to multiply or divide two complex numbers, it is much easier to write them in the form Aei
θ
and multiply/divide the A’s while adding or subtracting the phases.
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PHYS 3100
Because we know the functional form of an AC driving voltage or current (i.e., some constant
multiplied by cos(ωt+φ)), we can see that anything requiring us to take the first time derivative
of that function will give -ω sin(ωt+φ). We could write this with the cosine function instead,
using the fact that cos(θ-90°) = cos(θ-π/2) = sin(θ).
Specifically, if we have I(t) = Im cos(ωt + θ), and we know that V = L dI/dt for an inductor, we
could find
(
V ( t ) = L I m ( − ω ) cos ω t + θ − 90 
)
Of course, since our original current I(t) could also have been written as Imejθ (as always,
omitting the e jωt) we could also write this as
V ( t ) = I m ( − ω ) L e j θ e− j π
/2
However, you can use Euler’s formula to quickly see that e-j π/2 = -j, leaving us with
V ( t ) = j ω L I m e jθ = j ω L I ( t )
This leads us to the frequency domain. Now, when we see d/dt, we can replace it with jω. Of
course, this would have been clear from the first place if we had just written I(t) = Imejθ e jωt. This
is another benefit of the Euler notation. In angle notation, we can write V = ω L Im ∠(θ + 90°).
For the case of the capacitor, where the relation is instead I = C dV/dt, the same procedure gives
us I(t) = j ω C V(t), which we can solve for V to get
V (t) =
(
I
1
I ( t ) = m ∠ θ − 90 
jω C
ω C
)
This notation makes it clear that the voltage leads the current by 90° in an inductor and it lags the
current by 90° in a capacitor. Since the current and voltage are in phase in a resistor, we just get
the familiar V(t) = I(t) R for it.
As you first saw in introductory physics, a useful concept in a non-DC circuit is the impedance
of the circuit. This is a complex combination of the resistance (the real part, from the resistors)
and the reactance (the imaginary part, from any capacitors/inductors). We can then say that the
reactance of an inductor is XL = ω L and the reactance of a capacitor is XC = -1/(ω C). The
complex impedance is just R for a resistor, and j XL or j XC for the inductor or capacitor.
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PHYS 3100
Series and Parallel RLC Circuits
If we have a circuit consisting of multiple elements in parallel with one another, the big idea is
this: the voltage is the same across all elements. On the other hand, if we have a series circuit,
the big idea is that the current is the same throughout the circuit. Remembering these two
things will carry us far.
In introductory physics, you typically only see series RLC circuits. In a series circuit, the
individual impedances add to produce the final (total) impedance. Since XL and XC work in
opposite directions, their difference gives the total reactive impedance and that is combined with
the resistive impedance to get the magnitude of the total in the same way as you have seen earlier
Z=
R2 + ( X L − X C )
2
If we kept this in complex form, we would write Z = R + i (XL-XC). Since our resistances are
real, our inductive reactances are imaginary and positive, and our capacitive reactances are real
and negative, we can see that we get the expression above for Z by just adding all impedances
together.
On the other hand, parallel impedances are combined using the same formula used for resistors
in parallel:
1 1
1
=
+
+
Z || Z 1 Z 2
For example, assume we are given a parallel circuit consisting of a 35 Ω resistor, a 50 mH
inductor, and a 670 µF capacitor being driven by 180 V oscillating at 60 Hz. First, find XL and
XC using the formulas above. We get XL = 2 π f L = 18.85 Ω ∠ 90° and XC = 1 / (2 π f C) = 3.96
Ω ∠ -90°. The total impedance is then
1
1
1
1
=
+
+
Z 35 Ω + 0 i 0 + i 18.85 Ω
0 − i 3.96 Ω
giving us Z = 0.7036 – i 4.9124 Ω. Since the voltage across each element is the same, we get the
following currents:
180 V ∠ 0 
IR =
= 5.143 A ∠ 0 

35 Ω ∠ 0
Notice that the voltage and current are in phase in the resistor. In the inductor, we get
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PHYS 3100
180 V ∠ 0

IL =
=
9
.
549
A
∠
−
90
18.85 Ω ∠ 90
The current is 90° behind the voltage in the inductor. In the capacitor, we get
180V ∠ 0
IC =
= 45.45 A ∠ 90

3.96 Ω ∠ − 90
As expected, the current is 90° ahead of the voltage in the capacitor. This gives a total current of
5.143 A + i 35.901 A. Notice that we get the same current (verifying Kirchoff’s current law) if
we take the voltage and divide it by the total impedance
180V ∠ 0
IT =
= 36.268 A ∠ 81.85

4.963 Ω ∠ − 81.85
If we connect the same elements in a series circuit, we combine the impedances (which are
individually unchanged from the parallel case) to get 35 + i 14.89 Ω. The current is then
180 V ∠ 0
I=
= 4.732 A ∠ − 23.05

38.04 Ω ∠ 23.05
The voltage across each element is found by Ohm’s law, so we get
(
)(
)
VR = IR = 4.732 A ∠ − 23.05 35 Ω ∠ 0 = 165.62 V ∠ − 23.05
(
)(
)
VC = IX C = 4.732 A ∠ − 23.05 3.96 Ω ∠ − 90  = 18.74 V ∠ − 113.05
(
)(
)
VL = IX L = 4.732 A ∠ − 23.05 18.85 Ω ∠ 90 = 89.21V ∠ 66.95
If we add these three voltages together as complex quantities, we get 180 V ∠ 0°, as we would
expect from Kirchoff’s voltage law (which still works). Notice that the resistor’s voltage is in
phase with the current, that the capacitor’s voltage is 90° behind it, and that the inductor’s
voltage is 90° ahead of the current.
In comparing these two cases, we notice that the current is lower in the series circuit than in the
parallel circuit, as we would expect from our work with resistances. It’s also worth pointing out
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PHYS 3100
that we’ve gotten rid of a quirk from introductory physics. As you should remember, capacitors
in series were combined like resistors in parallel, and vice versa, to find the equivalent
capacitance. Inductors and resistors obey the same rules for their combinations, and the
capacitors were the “oddball” elements. All impedances we’ve seen work the same as resistors or
inductors, however. What has changed? Notice that the impedance of a resistor is just R and the
impedance of an inductor is proportional to L. For the capacitor, the impedance is proportional to
the inverse of C: XC = 1 / (2 π f C). This change has unified the behavior of all three kinds of
circuit elements, which makes our work a little simpler.
Impedance and Delta-to-Wye Transformations
Using the same ideas as in the case of pure resistance, we can find a ∆ to Y transformation for
impedance.
b
a
Zc
a
b
Zb
Z2
Z1
Za
Z3
c
c
The transformation to go from a ∆ to a Y is
Z1 =
Zb Zc
Z a + Zb + Zc
Z2 =
ZcZa
Za + Zb + Zc
Z3 =
Za Zb
Z a + Zb + Zc
while the Y to ∆ transformation is
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PHYS 3100
Za =
Z1 Z 2 + Z 2 Z 3 + Z1 Z 3
Z1
Zb =
Z1 Z 2 + Z 2 Z 3 + Z 1 Z 3
Z2
Zc =
Z 1 Z 2 + Z 2 Z 3 + Z1 Z 3
Z3
Source Transformations
Source transformations also work the same way with impedances that they did with pure
resistances. A voltage source V ∠ 0° in series with an impedance Z can be replaced by a current
source of magnitude (V ∠ 0°)/Z in parallel with the impedance Z. Similarly, a current source I ∠
0° in parallel with an impedance Z can be transformed to a voltage source (I ∠ 0°)Z in parallel
with the impedance Z.
We’ll do problem 9.41 from your book as an example. It is reproduced below.
The voltage of the sinusoidal source is given as 247.49 Cos(1000 t + 45°) V and we are asked to
find the Thevenin voltage and impedance at terminals a & b. First, we write the voltage in x + j y
form. The 45° angle means the real and imaginary components are equal in magnitude and are
both positive (since Sin(45°)=Cos(45°)), so we can write V = 175 + 175 j.
Next, we find the open-circuit voltage between a & b. To do that, we need to find the current in
the circuit, which requires us to find the impedance. The 100 Ω resistor is in series with the 100
mH inductor, so their impedances are additive. The frequency of the driving voltage is ω = 1000
so the inductor’s impedance is j1000*100mH = j 100 Ω, giving this branch a total impedance of
100 + 100 j. The capacitor has an impedance of 1/(jωC), giving us –j 100 Ω in that branch.
Combining the two parallel branches into an equivalent impedance gives us 1/[(1/(100+100j)) +
1/(-100 j)] = 100 – j 100. When we add this element to the inductor nearest the voltage source
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PHYS 3100
(which has the same j 100 Ω impedance as the other inductor) we get a final Zeff of 100 Ω. Our
current leaving the voltage source is then 175 + 175 j divided by Zeff, giving us 1.75 + 1.75 j A.
The voltage between points a and b is the same as the voltage across the capacitor or the voltage
across the combined inductor-resistor leg. We could work from the + terminal downstream to
find that voltage, or we could work from the – terminal up, which is what we’ll do. The voltage
we get is just equal to the current multiplied by the impedance equivalent to the two branches,
which we found earlier to be 100 – j 100 Ω. That leaves us with a Thevenin voltage of 350 + 0 j,
or 350 ∠ 0°.
Next, we need the short-circuit current. This is found more easily by recognizing that a short
circuit between a and b eliminates everything in the circuit except the first inductor and the
voltage source. The current would therefore be V/Z where Z is the first inductor’s impedance (j
100 Ω), so we get a short-circuit current of 1.75 – j 1.75. Finally, we find the Thevenin
impedance by using Z = VTh / Isc = (350)/(1.75 – j 1.75) = 100 + j 100. We could then redraw the
circuit with a voltage source of 350 ∠ 0° V connected in series with a 100 Ω resistor and a 10
mH (meaning XL = 100 j Ω) inductor.
The Node-Voltage Method
Unsurprisingly, the node-voltage method that worked so well in purely resistive DC circuits will
also work now. As an example, we’ll work problem 9-52, reproduced below.
As explained in the problem we want to use the node-voltage method to find the voltage across
the resistor V0. Our reference node will be the one at the bottom of the circuit (2 voltage sources
+ resistor). The voltages of the sources are given as
(
)
= 50 sin ( 2000 t − 16.26 ) V
V g1 = 20 cos 2000 t − 36.87  V
Vg 2

We need the voltage magnitudes and phases, but we need them using the cosine convention we
have adopted. For Vg1, we just get 20 ∠-36.87°. For Vg2, we need to use the fact that Sin(x) =
Cos(x-90°), allowing us to write Vg2 as 50 ∠-106.26°. Converting these to x + j y notation, we
would get
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PHYS 3100
Vg1 = 16 − j 12 V
Vg 2 = − 14 − j 48V
Now we’re prepared to apply the node-voltage method. The currents leaving the node in question
are (V0 – Vg1 )/ XL + V0/10 Ω + (V0 – Vg2 )/ XC. The impedance XL = j ω L = j 2 Ω and XC = 1 / (j
ω C) = -5 j Ω. Plugging in the numbers gives us
( 3.6 − 10.8 j ) = ( 0.1 −
0.3 j ) V0
so V0 = 36 ∠0° which we can rewrite as V = 36 cos(2000 t) V.
The Mesh Current Method
The mesh current method also works to find currents and voltages in AC circuits just as it does
for DC circuits. As an example, we’ll work problem 9-61 from your book (reproduced below).
If the voltage source Vg is 400 Cos(5000 t) volts, what is the voltage V0 across the 100 Ω
resistor? Using the mesh current method, we define two counterclockwise loops: the left loop’s
current is Ia and the right loop’s current is Ib. We need to find the impedances XL and XC. We get
XL = j ω L = 300 j Ω and XC = 1/(j ω C) = -100 j Ω. We can then write the following equations
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PHYS 3100
V g − I a X L − 50 ( I a − I b ) + 150 ( I a − I b ) = 0
− 150 ( I a − I b ) − 50 ( I b − I a ) + I b X C − 100 I b = 0
Solving these two equations for Ia and Ib (keeping in mind that Vg = 400 ∠ 0°), we get Ia = -0.8 j 1.6 and Ib = -1.6 + j 0.8. From there, finding the voltage V0 in the resistor just requires us to
find Ib R which is –160 + j 80. We can also write this as 178.89 ∠ 153.43° V.
Transformers
In the image below (adapted from an example in the All About Circuits PDF), we have a circuit
consisting of a driving AC voltage V1 which is powering a large inductor L1. Inductor L1 is
coupled to an inductor L2 in another circuit. The coupling provides the only source of power in
the second circuit (notice that the battery V2 is set at zero volts; it’s just there to make PSPICE
happy). The first circuit also contains a very small resistance (0.001 Ω) and a huge resistance (1
TΩ) connecting the two circuits, again for PSPICE. The only important elements in the second
circuit are the inductor L2 and a load resistance RL. Because of the magnetic coupling that exists
between the two inductors (specified in PSPICE with a “K” circuit element from the Analog
library), a changing current in L1 will produce a changing current in L2.
In this circuit, the voltages across the inductors are in phase with one another and with the
driving voltage. The current in L2 is also in phase with this voltage, but the current in L1 is 180°
out of phase with the voltage. The power consumed by RL is exactly the same as that consumed
by L1 (since we’ve made a perfect transformer) and the two oscillate together from 0 to their
maximum and back at twice the frequency of the AC driving voltage. The power consumed by
L2 also oscillates at the same frequency (twice the voltage source), but 180° out of phase with the
power in L1 and RL. Also, the power in (or consumed by) L2 is always negative, moving from 0
to a negative value which is equal to the negative of the power used in RL. This is reasonable,
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PHYS 3100
since the only real elements in the second circuit are RL and L2, so the power used by RL must be
supplied by L2.
A mesh-current approach to solving this circuit would include the internal impedance of the
source (Z1), the resistance R1 associated with the inductance L1 of the primary coil, the
impedance of the load ZL (RL in our example) and the inductance L2 and its associated resistance
R2 from the secondary coil. Our equations would be
V1 − ( Z1 + R1 + j ω L1 ) I1 + j ω M I 2 = 0
j ω M I1 − ( R2 + j ω L2 + Z L ) I 2 = 0
where M represents the mutual inductance between the two coils. As we would expect, it is a
symmetric term linking the EMF produced in the primary circuit to the current in the secondary
circuit and vice-versa. As we have done before, we choose I1 and I2 to both be clockwise
currents. Following your book, we introduce the simplifying notation
Z 11 = Z 1 + R1 + j ω L1
Z 22 = R L + j ω L2 + Z L
allowing us to write the currents as
I1 =
Z 22
V
2
2 1
Z 11 Z 22 + ω M
I2 =
jω M
jω M
V
=
I1
1
Z 22
Z 11 Z 22 + ω 2 M 2
The effective impedance seen by V1 can be found by dividing V1 by I1 which gives us Zint = Z11 +
(ω2M2/Z22). Subtracting out the internal impedance of the source, Z1, we get the impedance at the
terminals of the coil L1 to be
ω 2M2
Z term = R1 + j ω L1 +
R 2 + j ω L2 + Z L
The first two terms are easily seen to be just the impedance of the primary coil itself which can
be thought of as an ideal resistance in series with an ideal inductance. The third term is the effect
of the secondary circuit on the primary circuit, and is known as reflected impedance.
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PHYS 3100
Transformer Uses
The transformer’s purpose is to provide a way to increase voltages, currents or impedances from
one coil to the other. The transformer is very much like a lever in this sense: a lever serves as a
force multiplier so that you can exert a small force (say, 50 N) at one end and get a large force
(500 N, for example) at the other end. Because the lever has no way to provide energy, we know
that the price we pay for the force multiplication is a “distance division” since, in our example,
moving the 50 N end of the lever by 1 m will cause the 500 N end to move only 10 cm. The total
work done by the high-force end is exactly the same (neglecting friction, etc.) as the total work
done on the low-force end.
Transformers are similar in that they can be used as voltage multipliers, enabling a 12 V signal to
be stepped up to 240 V, for example. The trade-off is that the current on the 240 V side will be
only 1/20 of the current on the 12 V side. We can reverse the connections and make a step-down
transformer which would take the 12 V signal and make it 12/20 = 0.6 V, although that end of
the transformer could now supply 20 times the current of the 12 V end.
The transformer can also be considered to be an impedance multiplier/divider. A nice example of
this is found in section 9.7 of the “All About Circuits” reference, where two different 1000 W
heating elements are examined. One is designed for use with a 250 V circuit while the other is
made for a 125 V power supply. By Ohm’s Law, we can see that the high-voltage heating
element must have a resistance of 2502/1000 or 62.5 Ω while the low-voltage element’s
resistance would be 1252/1000 or 15.625 Ω. Trying to use these elements in the wrong circuits
would not work well. We would generate only 1252/62.5 Ω = 250 W of heat in one case, and
2502/15.625 Ω = 4000 W (quite probably destroying it) in the other case.
We can imagine using a transformer to convert the 125 V signal to 250 V, or vice versa, as
needed. From Faraday’s law of induction, we remember that the EMF induced when the
magnetic flux through an N-loop coil changes at a rate dΦ/dt is
ξ =− N
dΦ B
dt
If we have two coils and they are linked perfectly (which is not really an attainable goal, but we
can get very close to perfect linkage by wrapping each coil around a single iron core since iron is
extremely good at directing and confining magnetic flux), they will share a common value of dΦ
/dt. That means, if we refer to one coil as the primary and the other as the secondary, we get
ξS
dΦ B ξ P
=−
=
NS
dt
NP
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PHYS 3100
or, equivalently, ξS NP = ξP NS. This tells us that the ratio of the voltages is equal to the ratio of
the number of turns in the coils. This makes sense if we think of each loop being a sort of
“pickup” coil: the more loops you have on the iron core, the higher the voltage.
Notice that we could just as easily have said that the two-to-one transformer needed to let us
swap heating elements in the previous example was really an impedance transformer. From the
point of view of the 125 V supply, it was looking out at a 15.625 Ω load when it was in reality
connected to a transformer wired to a 62.5 Ω load. The impedance was transformed by a factor
of four, while the voltage/current were only transformed by a factor of two. This is again
reasonable since the voltage and current change in opposite directions and their ratio is the
impedance. Twice the voltage at half the current gives V/I = Z = 4 times the impedance.
This is an important thing to consider in designing audio systems. It’s very important that the
speaker’s impedance is matched to the amplifier’s impedance. A mismatch can be fixed by
selection of the correct transformer. We already know that the maximum power is transferred
from a source to a load when the impedances are equal; the transformer lets that happen.
Steady-State AC Circuits
Once transient effects have dissipated in a circuit (something that frequently happens much faster
than human reflexes can detect), an RLC circuit will exhibit the properties of a damped (R),
driven (AC voltage source) harmonic oscillator (LC). One of the things we’ll want to be able to
find is the instantaneous power developed or consumed by a circuit element. This formula is the
same as it’s always been: P = I V. The difference now is that I and V are sinusoidally-oscillating
functions of time. The current in the circuit will oscillate with the same frequency as the driving
voltage, but not necessarily in phase with it. In general, we can write
V ( t ) = Vm cos ( ω t + θ V )
I ( t ) = I m cos( ω t + θ
I
)
By convention, we will define the origin of time t = 0 to be when the current is a maximum,
meaning our formulas become
V ( t ) = Vm cos ( ω t + θ V − θ I )
I ( t ) = I m cos( ω t )
Through the use of various trig identities, we can now write
p=
Vm I m
( cos(θ V − θ I ) + cos(θ V − θ I ) cos ( 2ω t ) − sin (θ V − θ I ) sin ( 2ω t ) )
2
Examining these three terms shows that the first does not oscillate in time, but rather provides a
constant baseline from which oscillations in power deviate. In the case of simple resistive AC
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PHYS 3100
circuits, as you saw in introductory physics, the phases θV and θI are equal and this term would
be cos 0° = 1. In that same case, the second term would be cos(2 ω t) and the third would
disappear. This gives a power that oscillates from a minimum of 0 (when cos(2 ω t) = -1) and a
maximum of VmIm (when cos(2 ω t) = +1). Notice that this oscillation in power has a frequency
of 2 ω, meaning the power oscillates twice for every oscillation of voltage or current.
Notice that a purely inductive or purely capacitive circuit will have phases θV and θI which differ
by 90°, meaning the first two terms are zero and the final one is ± sin(2 ω t) (positive for an
inductive circuit, negative for a capacitive one). Since this function will be negative half of the
time and positive the other half, the power consumed over one full cycle (or any number of
them) is zero. Negative power corresponds to the circuit returning power to the driving voltage;
this can only happen if there are reactive elements (inductors or capacitors) in a circuit, since the
power consumed by a resistor must be non-negative.
It is sometimes useful to break the above expression for power into a slightly different form:
p = P + P cos( 2 ω t ) − Q sin ( 2 ω t )
where P is known as average power or real power and Q is the reactive power. We can see
why if we go back to our earlier case of a purely resistive circuit: Q will be zero and we get that
the power p = P (1 + cos(2 ω t)) (which could also be written as p = 2 P cos2(ω t)). By inspection
of the previous formulas, we can see that
Vm I m
cos(θ V − θ I )
2
V I
Q = m m sin (θ V − θ I )
2
P=
As mentioned, Q will be positive in an inductive circuit (since θV > θI and the sine of an angle
between 0° and 90° is a positive number) and negative in a capacitive one (since θI > θV and the
sine of an angle between -90° and 0° is a negative number). The units of Q are known as VARs,
for Volt-Ampere-Reactive to distinguish them from the ordinary Watts used when describing
average or real power.
We can consider real power and reactive power to be two legs of a right triangle. The hypotenuse
is called apparent power and is the magnitude of what is known as the complex power S which
has units of Volt-Amperes, or VA.
The three kinds of power are connected to the three things we have measured in Ωs (resistance,
reactance, and impedance) as shown below
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PHYS 3100
Apparent Power
|S| = I*I Z
Reactive Power
Q = I*I X
φ
Real Power
P = I*I R
φ = Impedance
Phase Angle
There is an important notational change here. Instead of writing P = I2R as we have seen earlier,
we have to take into account the fact that the power may very well be a complex number. In this
case, we need to find the absolute magnitude of the complex power. If we write it as I = x + j y,
we can see that the magnitude of I will be √(x2+y2), or I2 = (x2+y2). We can get the same result if
we write I*I where the asterisk represents the operation of complex conjugation. We could
therefore write I*I = |I|2. Complex conjugation means we just change the sign of the imaginary
part of a number. This has no effect on the real part of a number, so we can write a few examples
as: 6* = 6 (-3 j)* = (+3 j) (-17 + 4 j)* = (-17 – 4j).
Similarly, we need the absolute magnitude of Z if this triangle is to work mathematically. Since
Z is in general complex and of the form R ± j X, we see that the Z*Z will again give us a purely
real quantity. The magnitude of Z is therefore √(Z*Z) = √(R2+X2) (which may be familiar from
your introductory physics course).
Returning to the power, I*I would give (x - j y)(x + j y) = x2 – j2y2 = x2 + y2. The triangle shows
that we need to keep this in mind when working with complex currents. It’s reasonably easy to
verify that if Z = R + j X, the apparent power will be I*I |Z|, the real power will be I*I R, and the
reactive power will be I*I X.
Apparent power is important, because it represents what current-carrying wires have to be
designed to carry. As you can see, though, apparent power will be larger than the real power (the
power that does work) because of the reactive component. A measure of this mismatch is known
as the power factor, which is the cosine of the angle φ shown above. This could also be written
as cos φ = P/S using basic trigonometry. We could write this in still another form as cos(θV - θI)
which was the first of our three terms describing the power in an AC circuit.
A circuit that is purely resistive (a toaster, for example) will have a power factor of one since it
has no reactive or capacitive components (to a good approximation). The impedance Z is just
equal to the resistance R, meaning XL = XC = 0 and the reactive power Q is therefore zero. When
the Q leg of our right triangle above disappears, we can see that S = P and our power factor = 1.
If a circuit is purely reactive, and there is no resistance, we know that Z = X and the real power
will be zero. In this case, S = Q and our power factor will be zero.
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PHYS 3100
What if we have a household appliance (perhaps with a large motor) plugged into an outlet
providing 120 Vrms AC at 60 Hz? If it can be represented as an inductance of 0.575 H in series
with a resistance of 125 Ω, what are the powers in this case? First, we can see that XL will be 2 π
(60 Hz) (0.575 H) or 216.8 Ω. From our earlier work, we could write this as 0 + j 216.8 Ω. The
resistance is just 125 Ω, and the two elements are in series, giving us Z = 125 Ω + j 216.8 Ω.
The voltage source can be represented as 120 V ∠ 0°. This gives an rms current Irms of 0.2395 –
j 0.4154 A or 0.4795 A ∠-60° . The voltage drop across the inductor will be 90.06 + j 51.93 V
or 103.96 V ∠ 30° (notice that it’s 90° ahead of the current) and 29.94 - j 51.93 V or 59.94 V ∠
-60° (in phase with the current) across the resistor.
The real power P can be expressed as either Irms* Irms R, which gives us 28.74 W, or Irms* Irms |Z|
Cos φ = 57.53 * Cos φ = 28.74 W since φ = Tan-1(X/R) = 60°. The reactive power would be
either Irms* Irms X = 49.839 VAR or Irms* Irms |Z| Sin φ = 57.53 * Sin φ = 49.839 VAR. We could
write the complex power as S = P + j Q = 28.74 + j 49.839 VA or Vrms Irms* or Irms* Irms Z (notice
that this is the complex Z, not its magnitude) which gives us the same thing. The magnitude of S,
or the apparent power, is then √(P2 + Q2) = Irms* Irms |Z| = S*S = 57.532 VA
As an example, the figure below illustrates problem 10-9 in your book. The current source
provides a current of 40 Cos 1250 t mA. Find the average, apparent, and reactive powers
absorbed by the load.
We can write the current as 0.040 ∠ 0° A (notice that this is Im, not Irms), and we calculate that
j XL = j 10,000 Ω and 1/(j ω C) = -j 10,000 Ω. The resistor and capacitor are in series, so their
impedances add to give 5000 Ω -j 10,000 Ω. That impedance is in parallel with the inductor’s
impedance of j 10,000 Ω, so the effective impedance of the whole circuit is 20,000 Ω + j 10,000
Ω. We can then calculate the voltage across the inductor (which is equal to the voltage across
the resistor-capacitor branch) by using V = I Z. We get a voltage of 800 + j 400 V.
To find the current in the branch containing the load, we divide the voltage by the impedance of
that branch to get (800 + j 400 V)/(5000 Ω - j 10000 Ω) = j 0.080 A (which we could write as
0.080 ∠90°).
The real power P = ½ I*I R (the ½ is present because we’re using Im instead of Irms) = 16 W. The
reactive power Q = ½ I*I X = -32 VAR, and the complex power S = P + j Q = ½ I*I Z = 16 – j
32 VA. The apparent power, or magnitude of S is 35.78 W.
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PHYS 3100
Maximum Power Transfer
In the case of a purely resistive circuit, we’ve already seen that the maximum amount of power is
transferred to the load when its resistance matches the internal resistance of the source. When a
circuit has a reactive component, things are similar but not identical. First, we reduce the circuit
as seen by the load to its Thevenin equivalent voltage in series with its Thevenin impedance. At
that point, the total impedance will be (RTh + RL) + j (XTh + XL) so the current (which is the same
in both the load and the equivalent impedance) will be
VTh
( RTh + RL ) + j ( X Th + X L )
I=
The average power consumed by the load is then I*I RL, or
(VTh * VTh ) RL
( RTh + RL ) 2 + ( X Th +
P=
XL)
2
If we want to find the values for RL and XL which maximize the average power delivered to the
load, we need to take the derivative of P with respect to these variables and set it equal to zero.
As shown in your book, this yields
∂ P − (VTh * VTh ) 2 RL ( X L + X Th )
=
=0
∂ X L ( R + R )2 + ( X + X )2 2
L
Th
L
Th
and
[
∂ P (VTh * VTh )
=
∂ RL
]
[ (R + R )
[( R + R
L
2
Th
L
Th
+ ( X L + X Th ) − 2 RL ( RL + RTh )
2
)2 + ( XL +
X Th )
]
2 2
]= 0
For the nontrivial solution (i.e., VTh and RL ≠ 0), we see from the first equation that the reactance
of the load must be the negative of the Thevenin equivalent reactance of the source: XL = -XTh.
Under that condition, the second equation requires that RL + RTh = 2 RL, so RL = RTh. Notice that
this reduces to ZL = ZTh*.
If Z is not freely adjustable to any value, the best results are to be had when XL is as close as
possible to –XTh and RL is as close as possible to √[RTh2 + (XL + XTh)2]. If the phase of Z is
restricted instead, maximum power will be transferred to the load when the magnitudes of ZTh
and ZL are equal.
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PHYS 3100
Bibliography
Electric Circuits, 7th edition, Nilsson & Riedel
Basic Circuit Analysis, 2nd edition, John O’Malley
Electric Circuits, 4th edition, Nahvi & Edminister
http://www.allaboutcircuits.com - Tony Kuphaldt
http://inst.eecs.berkeley.edu/~ee40/fa06/lectures/cch-LecturesF06.pdf - Connie Chang-Hasnain
Physics, 6th Edition, Cutnell & Johnson
Fundamentals of Physics, 7th Edition, Halliday, Resnick, & Walker
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